I am fairly new to programming and I am trying to convert a string containing a base 10 number to an integer value following this pseudo algorithm in c.

start with n = 0
read a character from the string and call it c 
if the value of c is between '0' and '9' (48 and 57):
   n = n * 10  (c-'0')
   read the next character from the string and repeat
else return n

here is the rough basics of what i wrote down however I am not clear on how to read a character from the string. i guess im asking if i understand the pseudocode correctly.

stoi(char *string){
  int n = 0;
  int i;
  char c;
  for (i = 0;i < n ; i  ){
      if (c[i] <= '9' && c[i] >= '0'){
          n = n *10  (c - '0')}
      else{
           return n
           }
  }
}

CodePudding user response：

Congrats on starting your C journey!

One of the most important aspects of strings in C is that, technically, there are none. A string is not a primitive type like in Java. You CAN'T do:

String myString = "Hello";

In C, each string is just an array of multiple characters. That means the word Hello is just the array of [H,e,l,l,o,\0]. Here, the \0 indicates the end of the word. This means you can easily access any character in a string by using indexes (like in a normal array):

char *myString = "Hello";
printf("%c", myString[0]); //Here %c indicates to print a character

This will then print H, since H is the first character in the string. I hope you can see how you can access the any character in the string.

CodePudding user response：

You were close, you just need to traverse the string to get the value of each digit.

Basically you have two ways to do it.

Using array notation:

int stoi(const char *str)
{
    int n = 0;

    for (int i = 0; str[i] != '\0'; i  )
    {
        char c = str[i];

        if ((c >= '0') && (c <= '9'))
        {
            n = n * 10   (c - '0');
        }
        else
        {
            break;
        }
    }
    return n;
}

or using pointer arithmetic:

int stoi(const char *str)
{
    int n = 0;

    while (*str != '\0')
    {
        char c = *str;

        if ((c >= '0') && (c <= '9'))
        {
            n = n * 10   (c - '0');
        }
        else
        {
            break;
        }
        str  ;
    }
    return n;
}

Note that in both cases we iterate until the null character '\0' (which is the one that marks the end of the string) is found.

Also, prefer const char *string over char *string when the function doesn't need to modify the string (like in this case).